# What is the integral of xe^x from -infinity to 0?

Jul 14, 2015

I found: ${\int}_{- \infty}^{0} x {e}^{x} \mathrm{dx} = - 1$

#### Explanation:

I would solve it by parts:
$\int x {e}^{x} \mathrm{dx} = x {e}^{x} - \int 1 \cdot {e}^{x} \mathrm{dx} = x {e}^{x} - {e}^{x} =$
$= {e}^{x} \left(x - 1\right) {|}_{-} {\infty}^{0}$$=$
$= - 1 - 0 = - 1$
[where I used the fact that: ${e}^{0} = 1 \mathmr{and} {e}^{-} \infty = \frac{1}{e} ^ \infty = 0$]